What happens to a liquid when it loses energy

Which way will the people on the right move? Why? Plz answer

Scrat [10]

Probably to the right because the force on the left is greater and will therefore overpower the 2N force and push to the right

3 0

2 years ago

A cylinder of radius R, length L, and mass M is released from rest on a slope inclined at angle θ. It is oriented to roll straig

inna [77]

Answer:

$\mu_s=\frac{1}{3}\tan \theta$

Explanation:

Let the minimum coefficient of static friction be $\mu_s$ .

Given:

Mass of the cylinder = $M$

Radius of the cylinder = $R$

Length of the cylinder = $L$

Angle of inclination = $\theta$

Initial velocity of the cylinder (Released from rest) = 0

Since, the cylinder is translating and rolling down the incline, it has both translational and rotational motion. So, we need to consider the effect of moment of Inertia also.

We know that, for a rolling object, torque acting on it is given as the product of moment of inertia and its angular acceleration. So,

$\tau =I\alpha$

Now, angular acceleration is given as:

$\alpha = \frac{a}{R}\\Where, a\rightarrow \textrm{linear acceleration of the cylinder}$

Also, moment of inertia for a cylinder is given as:

$I=\frac{MR^2}{2}$

Therefore, the torque acting on the cylinder can be rewritten as:

$\tau = \frac{MR^2}{2}\times \frac{a}{R}=\frac{MRa}{2}------ 1$

Consider the free body diagram of the cylinder on the incline. The forces acting along the incline are $mg\sin \theta\ and\ f$ . The net force acting along the incline is given as:

$F_{net}=Mg\sin \theta-f\\But,\ f=\mu_s N\\So, F_{net}=Mg\sin \theta -\mu_s N-------- 2$

Now, consider the forces acting perpendicular to the incline. As there is no motion in the perpendicular direction, net force is zero.

So, $N=Mg\cos \theta$

Plugging in $N=Mg\cos \theta$ in equation (2), we get

$F_{net}=Mg\sin \theta -\mu_s Mg\cos \theta\\F_{net}=Mg(\sin \theta-\mu_s \cos \theta)--------------3$

Now, as per Newton's second law,

$F_{net}=Ma\\Mg(\sin \theta-\mu_s \cos \theta)=Ma\\\therefore a=g(\sin \theta-\mu_s \cos \theta)------4$

Now, torque acting on the cylinder is provided by the frictional force and is given as the product of frictional force and radius of the cylinder.

$\tau=fR\\\frac{MRa}{2}=\mu_sMg\cos \theta\times R\\\\a=2\times \mu_sg\cos \theta\\\\But, a=g(\sin \theta-\mu_s \cos \theta)\\\\\therefore g(\sin \theta-\mu_s \cos \theta)=2\times \mu_sg\cos \theta\\\\\sin \theta-\mu_s \cos \theta=2\mu_s\cos \theta\\\\\sin \theta=2\mu_s\cos \theta+\mu_s\cos \theta\\\\\sin \theta=3\mu_s \cos \theta\\\\\mu_s=\frac{\sin \theta}{3\cos \theta}\\\\\mu_s=\frac{1}{3}\tan \theta............(\because \frac{\sin \theta}{\cos \theta}=\tan \theta)$

Therefore, the minimum coefficient of static friction needed for the cylinder to roll down without slipping is given as:

$\mu_s=\frac{1}{3}\tan \theta$

3 0

2 years ago

Read 2 more answers

Another name for a period is a family?

Zepler [3.9K]

If you are referring to science, then you are correct. A period is also known as a family.

8 0

2 years ago

With the switch closed, how does the voltage across the 20-ω resistor compare to the voltage across the 10-ω resistor?

jasenka [17]

In electricity, the most famous and basic equation is the Ohm's Law which relates the parameters voltage, current and resistance. One form of this law as written in equation is V = IR, where V is the voltage in volts, I is the current in amperes and R is the resistance in ohms. These parameters depends in the arrangements, whether it's series or parallel.

In a series connection, the voltage is greater across a high-resistance resistor. Therefore, the voltage is much greater for the 20-ohm resistor. However,if it is a parallel circuit, the voltage is just the same for both resistors.

5 0

3 years ago

A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the a

Whitepunk [10]

Answer:

C) one-half as great

Explanation:

We can calculate the acceleration of gravity in that planet, using the following kinematic equation:

$\Delta x=v_0t+\frac{gt^2}{2}$

In this case, the sphere starts from rest, so $v_0=0$ . Replacing the given values and solving for g':

$g'=\frac{2\Delta x}{t^2}\\g'=\frac{2(22m)}{(3s)^2}\\g'=4.89\frac{m}{s^2}$

The acceleration due to gravity near Earth's surface is $g=9.8\frac{m}{s^2}$ . So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.

5 0

2 years ago